[geocentrism] Zero gravity

From: Bernard Brauer <bbrauer777@xxxxxxxxx>

To: geocentrism@xxxxxxxxxxxxx

Date: Mon, 12 Mar 2007 14:07:46 -0700 (PDT)

A satellite company wanted to deliver it's satellite to a point
above the equator of the Earth where the effects of gravity
would be zero, so that...................IT WOULD NOT FALL!
So they hire a math scientist and he says 22,200 miles high.
So the satellite company hires a rocket company to deliver
the satellite to that level.
So what is all the fuss about satellites? Heliocentrists keep
asking, "why doesn't the satellite fall"? Because it's not supposed to
at zero gravity. That was the whole point of putting it there. Sheesh!
The Earth itself "hangs on nothing" and the same
with the satellite - it "hangs on nothing."
Bernie
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Found on internet:
Question:
"I know that an object in geosynchronous orbit effectively "hovers" over a
specific point on earth, due to it actually orbiting the planet at the same
relative speed as the earth is revolving.
Yet the object does not actually "fall". So how high up does an object need to
be before it can do this without 'falling'?"
Answer:
So how do we determine this velocity and height?
F=ma
Ft (tangential force, the component of our force causing acceleration required
for geosynchronous orbit) = R*alpha (the radius from the point mass center of
earth's gravity multiplied with the angular acceleration {the rate at which the
object speeds up or slows down})
If you followed all that, then you should guess that for our geosynchronous
orbit the "alpha" must be zero (the object doesn't really change speed as it
"hovers" over a point on earth), which means the tangential component of the
force is zero!
Fc (centripedal force, the component of the force that causes the object to
constantly change direction, i.e., "fall" towards earth at just the correct
rate that it keeps falling downward at just the right speed that as it hurtles
forward it never really hits the earth) = m*v^2/R (the object's mass times it's
forward velocity squared divided by the radius at which it's orbiting).
But what *causes* this m*v^2/R force?
F=G*m1*m2/R^2 (or in intuitive terms, the force due to gravity is proportional
to the masses of the objects in question and inversely proportional the the
square of the distance between their centers of gravity).
So we'll call m2 Earth, and replace it with Me and we already refered to the
mass of the object as m, so m1=m.
So the duelling Newtonian determinations of the force on the object can be set
on separate sides of the equation:
m*v^2/R = G*m*Me/R^2
And calling in for algebraic backup, we can simplify things to:
v = sqrt(G*Me/R)
G is a constant and Me is something that we can look up in a Physics book. So
what we've got is a generic equation relation between v and R. What we haven't
yet used in our equation is the fact that we're specifically after
geosynchronous motion.
Or the v=R*omega (velocity at some radius is equal to the radius times the
rotational speed).
omega is one we can figure out ourselves, as it turns out the earth rotates
once every... (wait for it)... 23 hours and 56 minutes! Okay, just kidding.
Once every 24 hours. In terms friendly to calculations that means one
revolution (2*pi radians) per (24 hours/day * 60 minutes/hour * 60
seconds/minute) 86400 seconds.
So now we've got two equations with two unknowns which we can solve a couple of
different ways. I'm not going to go through the bother of describing the actual
mechanics (the process is to set one equation such that one unknown is isolated
and then plug that equation into the other), but what you end up with is
v = sqrt(G*Me*omega/v)
v^2 = G*Me*omega/v
v^3 = G*Me*omega
v = [(6.67 x 10^-11 N*m^2/kg^2) * (5.98 x 10^24 kg) * (2*pi/86400 s)]^1/3
v = 3070 m/s = 6870 mph
we can then put that into either of our equations to determine R:
R = v/omega = 42.3 x 10^6 m (the distance from the center of the earth) or,
subtracting off the radius of the earth (say at the equator),
d = R - Re = 35.8 x 10^6 m = 35800 km or 22200 miles above the equator.
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